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The following figure shows the data of students enrolled in $5$ years $(2014\;\text{to}\; 2018)$ for two schools $P$ and $Q$. During this period, the ratio of the average number of the students enrolled in school $P$ to the average of the difference of the number of students enrolled in schools $P$ and $Q$ is _______.

  1. $8 : 23$
  2. $23 : 8$
  3. $23 : 31$
  4. $31 : 23$
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Migrated from GO Electronics 2 years ago by Arjun

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We can calculate the number of students enrolled in both schools during the given period of time, 

  • The total number of students enrolled in school $P\text{(in 5 years)} = (3 + 5 + 5 + 6 + 4) \times 1000 = 23000$
  • Average number of students enrolled in school $P = \frac{23000}{5}$
  • And, the total number of students enrolled in school $Q\text{(in 5 years)} = (4 + 7 + 8 + 7 + 5) \times 1000 = 31000$
  • Average number of students enrolled in school $Q = \frac{31000}{5}$
  • Average of the difference of the number of students enrolled in school $P$ and $Q = \frac{31000}{5} – \frac{23000}{5} = \frac{8000}{5}$

Now the required ratio $ = \dfrac{\frac{23000}{8}}{\frac{8000}{5}} = \frac{23}{8}.$

$\therefore$ The ratio of the average number of the students enrolled in school $P$ to the average of the difference of the number of students enrolled in schools $P$ and $Q= 23:8.$


$\textbf{Short Method:}$  While calculating the ratio, common entities are cancelled.

  • The total number of students enrolled in school $P=3 + 5+ 5 + 6 + 4 = 23$
  • The total number of students enrolled in school $Q=4 + 7 + 8  + 7  + 5  = 31$

The difference between the number of students enrolled in school $P$ and school $Q=31 - 23 = 8$

Now the required ratio $ = 23:8.$

So, the correct answer is $(B).$

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✌ Edit necessary (RANJEETKUMBHAR “in ration calculation it should be: (23000/5)/(8000/5)”)
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Total number of students enrolled in school P:   $3K+5K+5K+6K+4K=23K$

Total number of students enrolled in school Q:  $4K+7K+8K+7K+5K=31K$

Difference between the number of student enrolled in school P & school Q :  $31K-23K=8K$

Required ratio :  $\frac{23K}{8K}$

Option B is correct.
Answer:

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